newtons second law of motion which recall from

Question

Newton's Second Law of motion, which recall from the earlier section that can be written as:

m(dv/dt) = F (t,v)

Here F(t,v) is the sum of forces which act on the object and may be a function of the time t and v that is the velocity of the object.  For our condition we will have two forces acting upon the object gravity, F_G = mg. acting in the downward direction and thus will be positive, and air resistance, F_A = -gv, acting in the upward direction and thus will be negative.  Placing all of this together in Newton's Second Law gives the subsequent as:

m(dv/dt) = mg - gv

To simplify the differential equation here we divide out the mass, m as

 (dv/dt) = g - (gv)/m   ..............................(1)

Here it is a first order linear differential equation, as solved, it will provide the velocity, v (in m/s) and of a falling object of mass m that has both air resistance and gravity acting on this.

So as to look at direction fields, this would be useful to have a few numbers for the different quantities in the differential equation. Conversely, let's suppose that we have a mass of 2 kg and that g = 0.392. Plugging it in (1) gives the subsequent differential equation.

dv/dt = 9.8 - 0.196v     .......................(2)

Here we take a geometric view of such differential equation. Let's assume that for several time, t, the velocity just occurs to be v = 30 m/s.  Notice that we are not saying as the velocity ever will be 30 m/s.  All that here we let's assume that by some possibility the velocity does occur to be 30 m/s at some time t. Thus, if the velocity does occur to be 30 m/s at some time t we can plug v = 30 in (2) to find.

dv/dt = 3.92

A positive derivative implies that the function in question, the velocity in such case, is rising, as if the velocity of such object is ever 30m/s for any time t the velocity should be rising at that time.